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Boundary condition, find E2

Short answer

Tangential E_x, E_z continuous: E_x2=3, E_z2=1. Normal D continuous: ε_r2 E_y2 = ε_r1 E_y1 → 5 E_y2 = 3×2 → E_y2=1.2. E₂ = 3 a_x + 1.2 a_y + a_z.

Use the boundary condition to find E2 in medium 2 with the boundary located at plane y = 0. Medium 1 is a perfect dielectric characterized by ε_r1 = 3, medium 2 is a perfect dielectric characterized by ε_r2 = 5, and the electric field in medium 1 is E1 = 3a_x + 2a_y + a_z.

Boundary at y = 0, so the normal is a_y and the tangential components are those along a_x and a_z. Medium 1: ε_r1 = 3. Medium 2: ε_r2 = 5. E₁ = 3 a_x + 2 a_y + a_z.

Boundary conditions for perfect dielectrics (no free surface charge):

  • Tangential E is continuous: E_t1 = E_t2.
  • Normal D is continuous: ε_r1 E_N1 = ε_r2 E_N2.
Tangential:  E_x2 = E_x1 = 3 ,  E_z2 = E_z1 = 1
Normal:      ε_r2 E_y2 = ε_r1 E_y1
             5 E_y2 = 3 × 2
             E_y2 = 6/5 = 1.2

Answer: E₂ = 3 a_x + 1.2 a_y + a_z (same units as E₁, e.g. V/m).

Check: D₁ = 3ε₀(3, 2, 1) has D_y = 6ε₀; D₂ = 5ε₀(3, 1.2, 1) has D_y = 6ε₀, so the normal flux density is continuous.

✓ Answer checked by Claude Opus 5.5 (high effort)

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